some confusion in mathematical induction I have  some  confusion in mathematical  induction(Picture Given below). I  don't  know  how it come ?

My attempt:
$a_1 = b_1 + 1, a_2 = b_2+ 1,............................,a_n= b_n+1$
then  $a_1 + a_2 + a_3 +.......+a_n =  b_1 + b_2 +.......+ b_n  + (1+.....+1)= b_1 + b_2 +.......+ b_n  + n$
Im confusing in the given  above answer  (mark in red box)  How it come ?
 A: Well, $$
a_{n-1} =b_n - 1 \\
a_{n} = b_{n} +1 
$$
from here we get $a_{n} - a_{n+1} = 2$ by subtraction, a relation involving only the $a_n$. It can be proved by induction now that $a_{n} = a_0 +2n$. The base case and induction case are both really obvious.
But then $b_n = a_n - 1$ so from the expression of $a_n$ we get $b_n = (2n-1) + a_0$.
A: Since $b_n+1=a_n$ and $a_{n-1}=b_n-1$, it follows that $a_n-2=a_{n-1}$.  
Now can you prove by induction that $a_n=2n+a_0$?
Now, from the first equation I wrote, $b_n=a_n-1=a_0+2n-1=2n-1+a_0$.
A: We wan to prove $\color{blue}{a_n = 2n +a_0}$ and $\color{blue}{b_n = 2n-1+a_0}$
Base step: $n= 0$
$\color{blue}{a_0 = 2*0 + a_0}$ and $b_0 +1 = a_0$ so $\color{blue}{b_0 = 2*0 - 1+ a_0}$.
Induction case: If we assume it is true for $n=k$
If $a_k = 2*k + a_0$ and $b_k = 2*k - 1+ a_0$ then
$2*k+ a_0 = a_k = a_{(k+1)-1} = b_{k+1} - 1 $ so $\color{blue}{b_{k+1} }=a_0 + 2k+1 \color{blue}{= 2(k+1) - 1+a_0}$.
And $\color{blue}{a_{k+1}} =b_{k+1} + 1\color{blue}{ = 2(k+1) + a_0}$
So it is true for $n = k+1$.
... 
That's it.  A proof by induction.
